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(B) Given that $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a}+\vec{b}+\vec{c}=\vec{0}$.Since $\vec{a}+\vec{b}+\vec{c}=\vec{0}$,we hav

Vedclass · Accepted Answer

$\vec{a} 	imes \vec{b}=\vec{b} 	imes \vec{c}=\vec{c} 	imes \vec{a} 
eq \vec{0}$

Vedclass · Answer

(B) Given that $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a}+\vec{b}+\vec{c}=\vec{0}$.Since $\vec{a}+\vec{b}+\vec{c}=\vec{0}$,we have $\vec{a}+\vec{b}=-\vec{c}$.Taking the cross product with $\vec{a}$ on both sides: $\vec{a} 	imes (\vec{a}+\vec{b}) = \vec{a} 	imes (-\vec{c})$.This gives $\vec{a} 	imes \vec{a} + \vec{a} 	imes \vec{b} = -\vec{a} 	imes \vec{c}$,which simplifies to $\vec{0} + \vec{a} 	imes \vec{b} = \vec{c} 	imes \vec{a}$.So,$\vec{a} 	imes \vec{b} = \vec{c} 	imes \vec{a}$.Similarly,taking the cross product with $\vec{b}$ on both sides: $\vec{b} 	imes (\vec{a}+\vec{b}) = \vec{b} 	imes (-\vec{c})$.This gives $\vec{b} 	imes \vec{a} + \vec{b} 	imes \vec{b} = -\vec{b} 	imes \vec{c}$,which simplifies to $\vec{b} 	imes \vec{a} = \vec{c} 	imes \vec{b}$,or $\vec{a} 	imes \vec{b} = \vec{b} 	imes \vec{c}$.Thus,$\vec{a} 	imes \vec{b} = \vec{b} 	imes \vec{c} = \vec{c} 	imes \vec{a}$.Since $\vec{a}, \vec{b}, \vec{c}$ are unit vectors forming an equilateral triangle,the cross products are non-zero. Therefore,$\vec{a} 	imes \vec{b} = \vec{b} 	imes \vec{c} = \vec{c} 	imes \vec{a} 
eq \vec{0}$.

Let $\vec{a}, \vec{b}, \vec{c}$ be unit vectors such that $\vec{a}+\vec{b}+\vec{c}=\vec{0}$. Which one of the following is correct?

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